Question

How do you solve the exponential equation `2^(x+1)=16^(x+2)`?

Answer 1

Real solution:

`x = -7/3`

Complex solutions:

`x = -7/3+(2kpii)/(3 ln 2)" "` for any `k in ZZ`

Explanation:

If `a > 0` then for any numbers `b, c` we have:

`(a^b)^c = a^(bc)`

So:

`2^(x+1) = 16^(x+2) = (2^4)^(x+2) = 2^(4(x+2))`

The function `f(x) = 2^x` from `RR` to `(0, oo)` is strictly monotonic increasing and therefore one to one. So if `a, b` are Real then:

`2^a = 2^b" "<=>" "a = b`

So in our example, we must have:

`x+1 = 4(x+2) = 4x+8`

Subtract `x+8` from both sides to find:

`-7 = 3x`

Divide both sides by `3` and transpose to get:

`x = -7/3`

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Complex solutions

Note that:

`e^(2kpii) = 1`

for any integer value of `k`

Hence:

`2^((2kpii)/ln 2) = e^(ln 2 * ((2kpii)/ln 2)) = e^(2kpii) = 1`

Hence for Complex values of `a, b` we have:

`2^a = 2^b" "<=>" "a = b+(2kpii)/ln 2 " for some " k in ZZ`

So in our example we must have:

`4x + 8 = x + 1 + (2kpii)/ln 2`

Subtract `x+8` from both sides to get:

`3x = -7+(2kpii)/ln 2`

Divide both sides by `3` to get:

`x = -7/3+(2kpii)/(3 ln 2)`